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In coordinates, write a differential form as $\eta ={\sum }_I{\eta }_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$. Its exterior derivative isFor $r=0$ this is the ordinary differential $df$. The chain rule, together with cancellation of symmetric second derivatives against the exterior product, proves that these formulas agree under coordinate changes. They give an $\mathbb{R}$-linear map $d:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ with $d^2=0$ and $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^r\alpha \wedge d\beta$ when $\alpha$ has degree $r$.

For a differential one-form $\omega ={\sum }_j{\omega }_{j}d{x}^j$, evaluation givesExpanding $X(\omega (Y))-Y(\omega (X))$ gives these coefficient derivatives plus ${\sum }_j{\omega }_j{\sum }_i(X^i{\partial }_i{Y}^j-Y^i{\partial }_i{X}^j)$. The latter is $\omega ([X,Y])$ by the Lie bracket of vector fields formula. Hence

A connection on a vector bundle is an $\mathbb{R}$-linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $D(f\sigma )=df\otimes \sigma +fD\sigma$. Evaluation on a vector field gives $D_X\sigma$, which is $C^{\infty }$-linear in $X$ and satisfies $D_X(f\sigma )=X(f)\sigma +f{D}_X\sigma$. The covariant exterior derivative extends this to bundle-valued forms byThis respects $(f\alpha )\otimes \sigma =\alpha \otimes (f\sigma )$: applying the rule on either side gives the same extra term $df\wedge \alpha \otimes \sigma$. It therefore defines an operator on bundle-valued forms independently of their local tensor representations. In a local frame with connection matrix $A$, its coefficient-column formula is $d^E=d+A\wedge$.

Define the curvature form of a connection by $R\sigma =(d^E{)}^2\sigma$. The extra terms in $(d^E{)}^2(f\sigma )$ cancel, so this equals $f(d^E{)}^2\sigma$. Equivalently the local formula iswhich acts fibrewise on the section coefficients. Thus curvature is a smooth endomorphism-valued two-form, a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(E)$.

For an $E$-valued one-form $a$, applying the ordinary one-form formula in a frame givesTaking $a=D\sigma$ proves the required curvature identityThe bracket correction is what makes the right side tensorial in the vector fields and the section.

Now define the endomorphism bundle connection by $({\tilde{D}}_X\alpha )(s)=D_X(\alpha s)-\alpha (D_{X}s)$. This expression is $C^{\infty }$-linear in $s$: the two terms involving $X(f)\alpha (s)$ cancel when $s$ is replaced by $fs$. It is therefore a smooth bundle endomorphism. It is $C^{\infty }$-linear in $X$, andThese identities prove it is a connection on a vector bundle. In a local frame it reads ${\tilde{D}}_X\alpha =X(\alpha )+[A(X),\alpha ]$.

For the full commutator identity for an endomorphism connection, expand once more:Subtracting the expression with $X,Y$ interchanged cancels both middle terms. ThusHere $D_X,D_Y$ are differential operators on sections, and the square brackets are their operator commutators. In general they are not fibrewise bundle endomorphisms; the curvature operator becomes fibrewise linear after subtracting $D_{[X,Y]}$.

Subtracting ${\tilde{D}}_{[X,Y]}\alpha$ from this identity gives the curvature of an endomorphism bundle connection:Consequently $\tilde{R}=0$ precisely when every $R_p(X,Y)$ commutes with every endomorphism of $E_p$. Arbitrary such endomorphisms can be extended locally in a frame, so this is a fibrewise assertion. A matrix commuting with all diagonal matrix units is diagonal; commuting also with all off-diagonal matrix units forces all its diagonal entries to coincide. The centre of the full endomorphism algebra therefore consists of scalar matrices. For rank $k>0$, define the smooth two-form $\omega =k^{-1}\mathrm{tr}R$. We obtain the scalar-curvature criterion for a flat endomorphism connectionThe converse follows immediately because scalar endomorphisms commute with all endomorphisms. For the zero bundle both curvatures vanish and one can take $\omega =0$. Scalar curvature here means the scalar endomorphism-valued curvature of this bundle connection, not the scalar trace curvature of a Riemannian manifold.

Use product coordinates $x^i$ on $M$ and $y^{\alpha }$ on $N$. The lifted vector fields have the form $X={\sum }_i{X}^i(x){\partial }_{x^i}$ and $Y={\sum }_{\alpha }{Y}^{\alpha }(y){\partial }_{y^{\alpha }}$. The coefficients of each depend only on its own coordinate set, and the coordinate partial derivatives commute. The Lie bracket of vector fields formula therefore givesThus $\boxed{[X,Y]=0}$ for fields lifted separately from the two factors. The lift hypothesis matters: arbitrary product-manifold fields can have coefficients depending on both factors.

A left-invariant vector field $X$ on a Lie group satisfiesor equivalently $(L_g{)}_{\ast }X=X$ for every $g$. For $\xi \in T_{e}G$, defineThis is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(d{L}_g{)}_h{X}_h^{\xi }=X_{gh}^{\xi }$, so $X^{\xi }$ is a left-invariant vector field. Conversely, left invariance forces any such field to have this form with $\xi =X_e$. Evaluation at $e$ and the construction above are inverse linear maps. HenceThe Lie bracket of vector fields is their commutator as derivations on smooth functions: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a diffeomorphism: conjugating the derivation operators by pullback of a smooth function preserves their commutator. Thereforefor left-invariant $X,Y$. These fields are consequently closed under the bracket.

Transport the bracket to the tangent space byBilinearity and antisymmetry come from the commutator. The Jacobi identity follows by expanding the three nested commutators of derivation operators: their six terms cancel in pairs. Thus $T_{e}G$ is a Lie algebra, and the displayed correspondence is a Lie algebra isomorphism onto the algebra of left-invariant vector fields.

In Cartesian coordinates on Euclidean space, defineThis is an affine connection by the ordinary product rule. Its connection coefficients in this coordinate frame are all zero. The component formula for the Lie bracket of vector fields gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, so it is torsion-free. Also, since the Euclidean metric has constant coefficients,This is exactly metric compatibility. By uniqueness it is the Euclidean Levi-Civita connection. In particular constant vector fields are parallel, while the derivative of a general field is given by the boxed formula.

The geodesic flow of a closed negatively curved surface is a topologically transitive Anosov flow. In particular, its Jacobi-field hyperbolicity and the usual negative-curvature transitivity theorem allow the smooth Livsic theorem to be applied. Set $b(x,v)={\beta }_x(v,v)$. The given closed-geodesic integrals are exactly its periodic-orbit obstructions on $SM$. Therefore

We make the differential operators and signs explicit. On the unit tangent bundle, let $V$ generate positive fibre rotation, $X$ be the geodesic vector field and $H=[V,X]$. For the canonical coframe of a surface unit tangent bundle, writeand let $\omega$ be the Levi-Civita connection form with $\omega (V)=1$. This uses $\theta$ for a one-form, keeping $\beta$ for the given tensor. The dual frame is $(X,H,V)$, and the structure equations areThese follow by writing the two horizontal forms as a rotated local orthonormal coframe and $\omega =d\vartheta +{\omega }_0$, with ${\omega }_0$ its connection 1-form. Equivalently, differentiating the sine and cosine components of $X$ twice in the fibre angle givesThe curvature equation also gives $[X,H]=KV$. The Liouville volume of a surface geodesic flow will be taken positive in the base/positive-fibre orientation, namely $\mu =\alpha \wedge \theta \wedge \omega$.

In an oriented orthonormal frame at $x$, put $v=(\cos \vartheta ,\sin \vartheta )$. The symmetric tensor evaluation isSince $V={\partial }_{\vartheta }$, this proves $(V^3+4V)b=0$.

The Lie bracket of vector fields identities give an operator identity that supplies the required equation. First,Using $VH=HV-X$ and $XV=VX-H$, differentiate this expression once more to obtainFor $\phi =V^{2}u+u$ and $Xu=b$, the first term vanishes by the explicit fibre calculation. Hence

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

Using the coordinate formula for the Lie bracket of vector fields, the two components areThus $\boxed{[X,Y]=-(x^2/2){\partial }_x+xy{\partial }_y}$, which is not the zero vector field.

The integral curves of a vector field $X$ solve $\dot{x}=y$, $\dot{y}=0$, whereas those of $Y$ solve $\dot{x}=0$, $\dot{y}=x^2/2$. Their local flows areBoth formulas exist for all real parameters, so both are complete vector fields. Direct composition givesAt $(1,0)$ their first coordinates differ by $ts/2$. Hence these local flows do not commute when $s,t$ are nonzero.

For the Lie bracket of vector fields, start at $(2,0)$. An integral curve of a vector field $[X,Y]$ isIt solves $\dot{x}=-x^2/2$, $\dot{y}=xy$, but $x(t)$ diverges as $t\downarrow -1$. It cannot extend through that finite time as a curve in ${\mathbb{R}}^2$. Complete vector fields are therefore not closed under the Lie bracket.

For the sum $X+Y$, its integral curves of a vector field obey $\dot{x}=y$, $\dot{y}=x^2/2$. Substitution of $x=(1+\alpha t{)}^{-2}$ and $y=\beta (1+\alpha t{)}^{-3}$ requiresChoosing $\alpha =-1/(2\sqrt{3})$ gives the integral curve of a vector fieldIt starts at $(1,1/\sqrt{3})$ and escapes to infinity at the finite time $2\sqrt{3}$. Complete vector fields are not closed under addition either. Both failures use the same pair of complete vector fields.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

Write the differential one-form as $\omega ={\sum }_i{\omega }_{i}d{x}^i$, and the vector fields as $X=X^i{\partial }_i$, $Y=Y^i{\partial }_i$. The coordinate definition of the exterior derivative givesExpand the right side of the invariant formula:The last line is $\omega ([X,Y])$, by the coordinate formula for the Lie bracket of vector fields. Subtract it and interchange the two indices in the second sum. The remaining terms are exactly $d\omega (X,Y)$. ThereforeThis is the exterior derivative of a one-form evaluated on vector fields; it is valid in any frame, not just a commuting coordinate frame.

Let $\omega ={\theta }^{-1}(\alpha )$. The preceding invariant formula for the exterior derivative identifies the expression in the question withIts vanishing for every pair of global vector fields is equivalent to $d\omega =0$: global fields with arbitrary prescribed tangent values are available by bump-function extension. Thus it is exactly the condition that $\omega$ be a closed differential one-form.

By definition, the first de Rham cohomology is closed one-forms modulo exact differential forms. The hypothesis $H_{\mathrm{dR}}^1(M)=0$ makes every closed one-form exact. Hence $d\omega =0$ gives a globally smooth function $g$ with $\omega =dg$, andConversely, if $\alpha (X)=X(g)$, thenby the definition of the Lie bracket of vector fields. Equivalently, $d(dg)=0$. The bracket compatibility condition is necessary and, when $H_{\mathrm{dR}}^1(M)=0$, sufficient for a global potential. The potential is unique up to a constant on each connected component, since the difference of two potentials has zero differential.

The Lie algebra of a Lie group $G$ is the vector space $\mathfrak{g}=T_{e}G$ equipped with the bracket transferred from left-invariant vector fields. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_g{L}_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_{e}G$, defineThis is smooth and left-invariant, because $L_a{L}_g=L_{ag}$ and the chain rule applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thusis a vector-space isomorphism, whose inverse is evaluation at the identity.

The Lie bracket of vector fields is preserved by diffeomorphisms: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w{]}_e$. Bilinearity, antisymmetry and the Jacobi identity follow from the same properties of commutators of derivations. This supplies the asserted Lie algebra structure.

For the general linear group $G{L}_n(\mathbb{R})$, invertible matrices form an open subset of $M_n(\mathbb{R})$, so $T_{I}G{L}_n(\mathbb{R})=M_n(\mathbb{R})$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_g{X}_A(H)=HA$. Therefore differentiating left-invariant matrix fields givesHenceThis is the general linear Lie algebra, with the ordinary matrix commutator; right invariance with the same identification would give the opposite sign.

Now let $F:G\to H$ be a smooth Lie group homomorphism. It sends the identity to the identity and induces the linear mapSince $F\circ L_g=L_{F(g)}\circ F$, differentiation givesThese fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,Apply this first with $w$ and then with $v$, and subtract the reversed order. The result isAt $e$, smooth functions detect tangent vectors, soThis proves that the differential of a Lie group homomorphism preserves Lie brackets, and hence that $F_{\ast }$ is a Lie algebra homomorphism.

For the exponential identity, let ${\eta }_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show ${\eta }_v(t+s)={\eta }_v(t){\eta }_v(s)$ for small times: translating the curve through $e$ gives the curve through ${\eta }_v(s)$. Repeating this identity extends the curve to all real times. This proves completeness of left-invariant vector fields without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives ${\eta }_v(t)={\eta }_{tv}(1)$. By the definition of the Exponential map of a Lie group, ${\exp }_G(v)={\eta }_v(1)$.

The field-related identity shows that $F({\eta }_v(t))$ is an integral curve of $X_{F_{\ast }v}$ starting at $e_H$. By uniqueness it equals ${\eta }_{F_{\ast }v}(t)$ for all times. At time one,This is the naturality of the Lie group exponential.

Apply this to the determinant homomorphism $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times }$. The derivative of the determinant at $I$ issince the permutation formula gives $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$. The left-invariant field on ${\mathbb{R}}^{\times }$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y^{\prime }=cy$, hence is $e^{ct}$. The matrix exponential determinant identity therefore follows from naturality:This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.

To identify it explicitly with the usual matrix exponential, the series $E(t)={\sum }_{k\ge 0}{t}^k{A}^k/k!$ converges in operator norm, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E^{\prime }=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $G{L}_n(\mathbb{R})$ and, by uniqueness, is the integral curve of $X_A$ through $I$. The matrix series is exactly the flow-defined exponential, justifying either notation in the determinant identity.

Use the standard symplectic form ${\omega }_0={\sum }_{j=1}^{n}d{q}_j\wedge d{p}_j$. A canonical transformation is a diffeomorphism $f$ preserving this symplectic form:Thus, writing $f(q,p)=(Q,P)$, the defining identity is ${\sum }_{j}d{Q}_j\wedge d{P}_j={\sum }_{j}d{q}_j\wedge d{p}_j$. For a local canonical transformation, require this on its open domain. In particular preservation of the symplectic form, rather than just of phase-space volume, is the condition in dimensions greater than two.

Throughout these solutions choose the Hamiltonian vector field convention ${\iota }_{X_F}\omega =dF$. It gives $\dot{q}=F_p$, $\dot{p}=-F_q$ and agrees with the positive coordinate Poisson bracket used below. The opposite convention for Hamiltonian vector fields changes signs in the Lie bracket of vector fields identity but none of the commuting or integrability conclusions.

A function $F$ is an integral of motion of $H$ when it is constant along every trajectory of the Hamiltonian flow. Infinitesimally,The antisymmetry of a Lie bracket of the Poisson bracket gives $X_{F}H=\{H,F\}=0$. Hence $H$ is also an integral of motion of $F$.

For the commuting assertion, ${\mathcal{L}}_{X_F}\omega =d({\iota }_{X_F}\omega )=d^{2}F=0$, because $\omega$ is closed. The contraction identity for a Lie bracket of vector fields then givesSince the symplectic form is a nondegenerate bilinear form at every point, the Hamiltonian vector fields commute:With the chosen convention the general identity is $[X_F,X_H]=X_{\{H,F\}}$. Equivalently, it follows by applying the Jacobi identity for the Poisson bracket to an arbitrary test function. Local flows commute whenever both compositions are defined; no global completeness assumption is needed.

The torsion tensor of an affine connection is $\mathcal{T}(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$, where the last term is the Lie bracket of vector fields. A torsion-free connection has $\mathcal{T}=0$. Since a coordinate basis commutes, this is equivalent there to ${\Gamma }^{\mu }{}_{\nu \rho }={\Gamma }^{\mu }{}_{\rho \nu }$; in a general basis its commutator must also be included.

The Levi-Civita connection is the unique torsion-free connection with metric compatibility, $\nabla g=0$. To derive its Christoffel symbols, lower the first index: ${\Gamma }_{\sigma \nu \rho }=g_{\sigma \mu }{\Gamma }^{\mu }{}_{\nu \rho }$. Metric compatibility givesWrite down this identity with the derivative indices $\rho ,\nu ,\sigma$, add the first two identities and subtract the third. Symmetry in the last two indices, supplied by the torsion-free connection, cancels all terms except twice ${\Gamma }_{\sigma \nu \rho }$. Raising the first index yieldsThis calculation proves uniqueness. Conversely, these Christoffel symbols are symmetric in $\nu ,\rho$ and satisfy $\nabla g=0$ by substitution, proving existence as well.

Describe the geodesic variation by a smooth map $F(s,\tau )$, with each curve $\tau \mapsto F(s,\tau )$ a timelike geodesic parametrized by proper time. Its tangent vector is $T=F_{\ast }{\partial }_{\tau }$ and its deviation vector is $S=F_{\ast }{\partial }_s$. Thus $S$ measures the first-order separation of neighboring members at the same value of proper time.

Mixed coordinate derivatives on the parameter surface commute. Consequently, along the variation,For an embedded two-dimensional variation this is the usual Lie bracket of vector fields on its image; more generally the same identity is read along the map $F$ using its commuting parameter derivatives. Since the Levi-Civita connection is torsion-free, it follows that ${\nabla }_{T}S={\nabla }_{S}T$.

For a smooth function, pullback of a smooth function is composition. The chain rule givesFor a vector field $Y$, use any local coordinates. To first order,Multiplying this inverse differential by $Y({\varphi }_s(x))$ givesThusThis is the Lie bracket of vector fields, whose action on a function is $X(Yf)-Y(Xf)$.

Write the covector field as $\omega ={\omega }_{\mu }d{x}^{\mu }$. From the contraction and product rule properties,Take $Y={\partial }_{\mu }$. The Lie bracket of vector fields is $[X,{\partial }_{\mu }]=-({\partial }_{\mu }{X}^{\nu }){\partial }_{\nu }$, soEquivalently, the covariant-factor contribution comes from ${\mathcal{L}}_{X}d{x}^{\mu }=d{X}^{\mu }=({\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$. The formula holds in any coordinate basis and is the one-form case of the coordinate tensor Lie derivative.